A Bernoulli differential equation is one of the form dy + P(x)y = Q(x)y" (*) dz y¹- 1-n transforms the Bernoulli equation into the linear equation du + (1 - n)P(x)u = (1 − n)Q(x). dz Observe that, if n = 0 or 1, the Bernoulli equation is linear. For other values of n, the substitution u = Consider the initial value problem zy' + y = 6zy², y(1) = −9. (a) This differential equation can be written in the form (*) with P(x)= |Q(x) = n = = (b) The substitution u = du + dx u = and will transform it into the linear equation (c) Using the substitution in part (b), we rewrite the initial condition in terms of x and u: |u(1) = (d) Now solve the linear equation in part (b), and find the solution that satisfies the initial condition in part (c). |u(x) = (e) Finally, solve for y. | y(x) =
A Bernoulli differential equation is one of the form dy + P(x)y = Q(x)y" (*) dz y¹- 1-n transforms the Bernoulli equation into the linear equation du + (1 - n)P(x)u = (1 − n)Q(x). dz Observe that, if n = 0 or 1, the Bernoulli equation is linear. For other values of n, the substitution u = Consider the initial value problem zy' + y = 6zy², y(1) = −9. (a) This differential equation can be written in the form (*) with P(x)= |Q(x) = n = = (b) The substitution u = du + dx u = and will transform it into the linear equation (c) Using the substitution in part (b), we rewrite the initial condition in terms of x and u: |u(1) = (d) Now solve the linear equation in part (b), and find the solution that satisfies the initial condition in part (c). |u(x) = (e) Finally, solve for y. | y(x) =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:A Bernoulli differential equation is one of the form
dy
+ P(x)y = Q(x)y"
(*)
dz
y¹-
1-n transforms the Bernoulli equation into the linear equation
du
+ (1 - n)P(x)u = (1 − n)Q(x).
dz
Observe that, if n = 0 or 1, the Bernoulli equation is linear. For other values of n, the substitution u =
Consider the initial value problem
zy' + y = 6zy², y(1) = −9.
(a) This differential equation can be written in the form (*) with
P(x)=
|Q(x) =
n =
=
(b) The substitution u =
du
+
dx
u =
and
will transform it into the linear equation
(c) Using the substitution in part (b), we rewrite the initial condition in terms of x and u:
|u(1) =
(d) Now solve the linear equation in part (b), and find the solution that satisfies the initial condition in part (c).
|u(x) =
(e) Finally, solve for y.
| y(x) =
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